Chapter 11, Problem 37CE

Instructions: You may use Excel, MegaStat, Minitab, JMP, or another computer package of your choice. Attach appropriate copies of the output or capture the screens, tables, and relevant graphs and include them in a written report. Try to state your conclusions succinctly in language that would be clear to a decision maker who is a nonstatistician. Exercises marked * are based on optional material. Answer the following questions, or those your instructor assigns.

1. a. Choose an appropriate ANOVA model. State the hypotheses to be tested.
2. b. Display the data visually (e.g., dot plots or line plots by factor). What do the displays show?
3. c. Do the ANOVA calculations using the computer.
4. d. State the decision rule for α = .05 and make the decision. Interpret the p-value.
5. e. In your judgment, are the observed differences in treatment means (if any) large enough to be of practical importance?
6. f. Given the nature of the data, would more data collection be practical?
7. g. Perform Tukey multiple comparison tests and discuss the results.
8. h. Perform a test for homogeneity of variances. Explain fully.

An ANOVA study was conducted to compare dental offices in five small towns. The response variable was the number of days each dental office was open last year. Research question: Is there a difference in the means among these five towns?

To determine

Choose appropriate ANOVA model. State the hypothesis.

Answer to Problem 37CE

One-factor ANOVA is used in this situation.

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5$

Alternative hypothesis:

$H_1:$ Not all means are equal.

Explanation of Solution

The given information is the mean number of days a dental office opened for five towns.

One-factor ANOVA is mainly used for comparing several columns of data. Therefore, the data comparison for mean productivity measurement of three plants can be model using one-factor ANOVA.

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5$

Mean number of days dental office opened for all five towns are equal.

Alternative hypothesis:

$H_1:$ Not all means are equal.

At least one pair of mean number of days is different.

To determine

Display the data for each plot and explain the plot.

Answer to Problem 37CE

The dot plot is given by:

Explanation of Solution

Calculation:

Dot plot:

Software procedure:

Step-by-step procedure for Dot plot using the Mega Stat is given below:

• • Open an EXCEL file.
• • In Mega Stat, select Analysis of variance and then One-Factor ANOVA.
• • In Input range drop down box, select Values.
• • Click on Plot the data.
• • Click OK.

The output using the Mega Stat software is given below:

Observations:

From the dot plot it is clear that mean number of days dental office in Chalmers opened is differ from others. Therefore, mean number of days dental office opened are significant.

To determine

Perform ANOVA.

Explanation of Solution

One-factor ANOVA and Turkey’s comparison:

Software procedure:

Step-by-step procedure for One-factor ANOVA and Turkey’s comparison using the MegaStat is given below:

• • Open an EXCEL file.
• • In Mega Stat, select Analysis of variance and then One-Factor ANOVA.
• • In Input range drop down box, select Values.
• • In post-hoc analysis select p less than 0.05
• • Click OK.

The output using the Mega Stat software is given below:

To determine

Explain the conclusion.

Interpret the p-value.

Answer to Problem 37CE

The means number of days dental office opened are significant at $\alpha =0.05$.

Explanation of Solution

Calculation:

From the MegaStat output, the p-value is 0.0019.

Rejection rule:

If $p\text{-value}\le \alpha$, reject $H_0$. Otherwise do not reject $H_0$.

Conclusion:

From the MegaStat output, the p-value is 0.0019.

That is less than the significance level $\alpha =0.05$.

That is, $p\text{-value}<\text{0}\text{.05}$.

Therefore, the null hypothesis is rejected.

Hence, means number of days dental office opened are significant.

Here, the p-values for output are very small. Therefore, samples were support the rejection of the null hypotheses.

To determine

Check whether there the observed difference in treatment means large enough to be of practical importance.

Answer to Problem 37CE

The observed difference in treatment means large enough to be of practical importance.

Explanation of Solution

From the dot plot or the individual value plot it clear that mean number of days dental office in Chalmers opened is differ from others. That is, the mean number of days dental office opened are different. Therefore, mean number of days dental office opened are significant.

Therefore, observed difference in treatment means large enough to be of practical importance.

To determine

Check whether more data collection is practical if nature of the data is given.

Explanation of Solution

Answer will vary. One of the possible answers is given below:

The samples are taken from the dental office in five towns. If the nature of the data is given then it is easy to collect samples from more dental office. Thus, the data collection is practical if nature of the data is given.

To determine

Perform Turkeys test using explain the results.

Answer to Problem 37CE

There is significant difference among mean number of times the dental office opened in Ulysses and Villa Nueve.

There is significant difference among mean number of times the dental office opened in Ulysses and Chalmers.

There is significant difference among mean number of times the dental office opened in Greenburg and Chalmers.

There is significant difference among mean number of times the dental office opened in Hazeltown and Chalmers.

Explanation of Solution

Calculation:

Significance level is $\alpha =0.05$.

 State the hypotheses:

Null hypothesis:

$H_0:{\mu }_i={\mu }_{j}i=1,2,3,4$

That is, all the pair treatment means are equal.

Alternative hypothesis:

$H_0:{\mu }_i\ne {\mu }_{j}i=1,2,3$

That is, at least one pair of treatment means differ.

Here, the Turkey’s test statistic is $T_{\text{calc}}=\frac{\left|\overline{y_i}-\overline{y_j}\right|}{\sqrt{\text{MSE}\left[\frac{1}{n_i}+\frac{1}{n_i}\right]}}$  where $\overline{y_i}$ and $\overline{y_j}$ are the mean of i^th and j^th group, MSE is the pooled variance of the c samples, and $n_i$ and $n_j$ are sample size of the i^th and j^th group.

Rejection rule:

If $p\text{-value}\le \alpha$, reject $H_0$. Otherwise do not reject $H_0$.

From the MegaStat output in part (c),

The value of the p-value for Ulysses and Greenburg is 0.3626.

The value of the p-value for Ulysses and Hazeltown is 0.0073.

The value of the p-value for Ulysses and Vila Nueve is 0.0.0073.

The value of the p-value for Ulysses and Chalmers is 0.0.0002.

The value of the p-value for Greenburg and Hazeltown is 0.8657.

The value of the p-value for Greenburg and Vila Nueve is 0.0833.

The value of the p-value for Greenburg and Chalmers is 0.0.0045.

The value of the p-value for Vila Nueve and Hazeltown is 0.0950.

The value of the p-value for Chalmers and Hazeltown is 0.0042.

The value of the p-value for Chalmers and Vila Nueve is 0.1414.

Conclusion:

For Ulysses and Greenburg:

The p-value is 0.3626, which greater than the significance level is $\alpha =0.05$.

That is, $p\text{-value}>\text{0}\text{.05}$. Therefore, the null hypothesis is not rejected.

Thus, there is no significant difference among mean number of times the dental office opened in Ulysses and Greenburg.

For Ulysses and Hazeltown:

The p-value is 0.2556, which greater than the significance level is $\alpha =0.05$.

That is, $p\text{-value}>\text{0}\text{.05}$. Therefore, the null hypothesis is not rejected.

Thus, there is no significant difference among mean number of times the dental office opened in Ulysses and Hazeltown.

For Ulysses and Vila Nueve:

The p-value is 0.0073, which is less than the significance level is $\alpha =0.05$.

That is, $p\text{-value}<\text{0}\text{.05}$. Therefore, the null hypothesis is rejected.

Thus, there is significant difference among mean number of times the dental office opened in Ulysses and Vila Nueve.

For Ulysses and Chalmers:

The p-value is 0.0002, which is less than the significance level is $\alpha =0.05$.

That is, $p\text{-value}<\text{0}\text{.05}$. Therefore, the null hypothesis is rejected.

Thus, there is significant difference among mean number of times the dental office opened in Ulysses and Chalmers.

For Hazeltown and Greenburg:

The p-value is 0.8657, which greater than the significance level is $\alpha =0.05$.

That is, $p\text{-value}>\text{0}\text{.05}$. Therefore, the null hypothesis is not rejected.

Thus, there is no significant difference among mean number of times the dental office opened in Hazeltown and Greenburg.

For Vila Nueve and Greenburg:

The p-value is 0.0833, which greater than the significance level is $\alpha =0.05$.

That is, $p\text{-value}>\text{0}\text{.05}$. Therefore, the null hypothesis is not rejected.

Thus, there is no significant difference among mean number of times the dental office opened in Vila Nueve and Greenburg.

For Greenburg and Chalmers:

The p-value is 0.0045, which is less than the significance level is $\alpha =0.05$.

That is, $p\text{-value}<\text{0}\text{.05}$. Therefore, the null hypothesis is rejected.

Thus, there is significant difference among mean number of times the dental office opened in Greenburg and Chalmers.

For Hazeltown and Vila Nueve:

The p-value is 0.0950, which greater than the significance level is $\alpha =0.05$.

That is, $p\text{-value}>\text{0}\text{.05}$. Therefore, the null hypothesis is not rejected.

Thus, there is no significant difference among mean number of times the dental office opened in Hazeltown and Vila Nueve.

For Hazeltown and Chalmers:

The p-value is 0.0042, which is less than the significance level is $\alpha =0.05$.

That is, $p\text{-value}<\text{0}\text{.05}$. Therefore, the null hypothesis is rejected.

Thus, there is significant difference among mean number of times the dental office opened in Hazeltown and Chalmers.

For Chalmers and Vila Nueve:

The p-value is 0.1414, which greater than the significance level is $\alpha =0.05$.

That is, $p\text{-value}>\text{0}\text{.05}$. Therefore, the null hypothesis is not rejected.

Thus, there is no significant difference among mean number of times the dental office opened in Chalmers and Vila Nueve.

To determine

Perform test of homogeneity of variance and explain the results.

Answer to Problem 37CE

There is enough evidence to conclude that variances are not different.

Explanation of Solution

Calculation:

State the hypotheses:

Null hypothesis:

$H_0:{\sigma }_1^2={\sigma }_2^2={\sigma }_3^2={\sigma }_4^2={\sigma }_5^2$

All the treatment variance are equal.

Alternative hypothesis:

$H_1:$ Not all ${\sigma }_j^2$ are equal.

At least one pair of the treatment variance differs.

Rejection rule:

If $H\ge H_{\text{Crtical}}$, reject $H_0$. Otherwise do not reject $H_0$.

Test statistics:

The Hartley’s test statistics H is given by:

$H=\frac{s_{\max }^2}{s_{\min }^2}$

Where, $s_{\max }^2$ and $s_{\min }^2$ is the maximum and minimum of sample variance respectively.

The sample variances are $s_1^2={10.83}^2,s_2^2={6.85}^2,s_3^2={11.17}^2,s_4^2={9.97}^2\text{and}{s}_5^2={9.42}^2$.

Here,

$\begin{array}{c}{s}_{\max }^2=\text{Max}\left({10.83}^2,{6.85}^2,{11.17}^2,{9.97}^2{9.42}^2\right)\\ ={11.17}^2\\ =124.7689\\ =124.77\end{array}$

$\begin{array}{c}{s}_{\min }^2=\text{Min}\left({10.83}^2,{6.85}^2,{11.17}^2,{9.97}^2{9.42}^2\right)\\ ={6.85}^2\\ =46.9225\\ =46.92\end{array}$

Substitute these values in the above equation.

Therefore,

$\begin{array}{c}H=\frac{124.77}{46.92}\\ =2.659\end{array}$

Thus, the Hartley’s test statistics is 2.659.

Degrees of freedom:

For between group:

$df\left(\text{Between group}\right)=c$,  where c is the number of groups.

Here, there are 3 groups. That is $c=5$. Substitute this value in the above formula.

Therefore,

$\begin{array}{c}df\left(\text{Between group}\right)=df\left(\text{Numerator}\right)\\ =d{f}_1\\ =5\end{array}$

Thus, between group degrees of freedom are 5.

For within group:

$df\left(\text{within group}\right)=\frac{n}{c}-1$, where c is number of groups and n is the sample size.

$\begin{array}{c}n=n_1+n_2+n_3+n_4+n_5\\ =6+4+6+5+5\\ =26\end{array}$

Substitute $c=5\text{ and }n=26$ in the above formula.

Therefore,

$\begin{array}{c}df\left(\text{within group}\right)=df\left(\text{denominator}\right)\\ =d{f}_2\\ =\frac{26}{5}-1\\ =5.2-1\\ =4.2\\ =4\end{array}$

Thus, within group degrees of freedom are 4.

Critical-value:

Procedure for the value of Hartley’s H using Table 11.5:

• • Go through the row corresponding to 5 in Table 11.5 of five percent critical values of Hartley’s $H=\frac{s_{\max }^2}{s_{\min }^2}$ .
• • Go through the row corresponding to 5 and column corresponding to the number of groups 4.
• • Obtain the value corresponding to (5, 4) from the table.

Thus, the critical-value is 25.2.

Conclusion:

Here, the H-test statistic is 2.659.

Here, $H\text{-statistic}\left(=2.659\right)<H_{\text{Critical}}\left(=25.2\right)$.

That is test statistic is less than the critical value.

Therefore, the null hypothesis is not rejected.

Thus, there is enough evidence to conclude that variances are not different.